Problem: $\overline{AC}$ is $3$ units long $\overline{BC}$ is $4$ units long $\overline{AB}$ is $5$ units long What is $\csc(\angle BAC)?$ $A$ $C$ $B$ $3$ $4$ $5$
Answer: $\csc(\angle BAC) = \dfrac{1}{\sin(\angle BAC)}$ How can we find $\sin(\angle BAC)$ SOH CAH TOA in = pposite over ypotenuse Opposite $= \overline{BC} = 4$ Hypotenuse $= \overline{AB} = 5$ $\sin(\angle BAC) = \dfrac{4}{5}$ $\csc(\angle BAC) = \dfrac{1}{\sin(\angle BAC)} = \dfrac{5}{4}$